How Does Charge Distribution Affect Electric Fields in Concentric Spheres?

[Arisa]

Homework Statement

A solid conducting sphere carrying charge q has radius a. It is inside a concentric hollow conducting sphere with inner radius b and outer radius c. The hollow sphere has not net chare. a) Derive expressions for the electric field magnitude in terms of the distance r from the center for the regions r<a, a<r<b, b<r<c, r>c.

Homework Equations

E = [1/(4*pi*epsilon_0)](q/r^2)

The Attempt at a Solution

I've managed to correctly answer the first two parts of the problem, however when it comes to b<r<c and r>c, I do not get the answers I should.
Apparently, for b<r<c, E = 0 since a -q cancels the inner +q. Then, for r>c, E = [1/(4*pi*epsilon_0)](q/r^2) since the total charge enclosed is +q again.

I think my problem lies in the fact that I don't fully comprehend what a concentric sphere is or how charge distribution on a concentric sphere works. Based on the solution, I feel I should intrepret that the neutral concentric sphere is neutral because it contain an equal number of positive and negative charges that have all collected on opposite surfaces - the negative charges on the inner surface of the concentric sphere (radius b) and the positive charges on its outer surface (radius c.) Otherwise, I don't quite understand how the -q and overall +q come into play...

Thank you very much for taking the time to read this!

 

[Astronuc]

This might help - http://hyperphysics.phy-astr.gsu.edu/hbase/electric/gaulaw.html

The inner sphere has charge +q. Being a conductor, the charge resides at the surface.

It would have a surface charge density of +q/4$\pi$ a².

This positive charge induces a corresponding -q charge on the inner surface of the hollow conductor (r=b), and consequently there is a +q charge on the outer surface r = c.

The electric flux is based on the enclosed charge, and the +q at r=a cancels the -q charge at r=b.

Then there is a charge +q at r=c.

The surface density of the charge at r = b is -q / 4$\pi$ b², and the surface charge density at r=c is +q / 4$\pi$ c².

 

[m2003]

Your understanding of the problem is correct. A concentric sphere is a sphere that shares the same center point with another sphere. In this case, the solid conducting sphere with charge q is inside the hollow conducting sphere with no net charge. This means that the total charge enclosed by the hollow sphere is zero, as you correctly pointed out.

For the region b<r<c, the electric field is indeed zero because the positive and negative charges on the inner and outer surfaces of the hollow sphere cancel each other out, resulting in no net electric field in this region.

For r>c, the electric field is given by the equation E = [1/(4*pi*epsilon_0)](q/r^2), as you mentioned. This is because the solid conducting sphere with charge q is now the only source of electric field in this region, and the total charge enclosed by the concentric spheres is still +q.

It is important to understand the concept of charge distribution on a concentric sphere in order to solve this problem correctly. The neutral concentric sphere contains an equal number of positive and negative charges, with the negative charges on the inner surface and the positive charges on the outer surface. This is because when a conductor is placed in an external electric field, the charges on the surface redistribute themselves in such a way that the electric field inside the conductor is zero.

I hope this helps clarify any confusion you had. Good job on solving the first two parts of the problem correctly, and keep up the good work!